Fourier Series

Chapter 3 Complex Fourier Series
Chapter 3.1 Complex Fourier Series exp(jωt) as a rotating vector
You always meet this complex exp(jωt)  function  in the electricity, automatics, acoustics…This is a f(t) special complex function  case.
The domain of the function is a real number and not a complex number as for an average complex function.
Other words. The complex number z=exp(jωt) is assigned to a real number time t. See Note
See animation. You return to article by typical windows pushbutton “Windows return arrow

Fig. 3-1
Complex function z=1exp(jωt) as a roatating vector ω=2.62/sec
Click  ANIMATION–>author Chetvorno
You see rotating black dot. Check the time of the 10 rotations, t=24sec–>T=2.4sec –>pulsation ω=2π/T=2.62/sec
The z=1exp(jωt) black dot location is just a complex function  in the complex numbers plane.
For example:
1exp(j0)=1+j0=1
1exp(j60º)=1exp(jπ/3)≈0.5+1+j0.8 
as Fig.3-1 (assume that ωt=π/3=60º)
You can use Euler formula 1exp(jωt)=sin(ωt)+jcos(ωt) too:
For example:
1exp(j0)=1+j0=1
1exp(jπ/2)≈0+j1=1j
Note
The complex function domain is generally a two-dimensional number, and the codomain is a two-dimensional number too. It means that we are in the four-dimensional space now! It’s difficult to imagine. Fortunately, many phenomenons are described by easy to imagine complex exp(jωt) function of time t.
More about complex numbers and functions–> see article in the upper tab.

Chapter 3.2  a*sin(ωt)+b*cos(ωt) formula as a complex version
The math operations of the trigonometric funtions are much easier when complex numbers are used
Examples
1. The sum trigonometric formula for a*sin(ωt)+b*cos(ωt) is complicated. But it’s very easy when complex numbers are used.
2. The sine functions differentiation and integration  is a simple  +/-90º vector rotation when complex numbers are used.

Chapter 3.2.1 c*exp(jωt) as a rotating vector
The coeffient c in the z=1*exp(jωt), is a real number c=1. Let’s consider any complex number, i.e. c=1.8+0.4j for example.

Fig. 3-2
c*exp(jωt) as a rotating vector
The real number 1 is rotating as a 1*exp(jωt).
The complex  number c=1.8+0.4j is rotating as a c*exp(jωt).
For example, you see  2 states of the rotating vectors 1*exp(jωt) and (1.8+0.4j)*exp(jωt).
When:
ωt=0º
ωt=ωt=π/3=60º

Chapter 3.2.2 a1cos(1ωt) i b1sin(1ωt) components as rotating vectors.
Let’s consider components a1cos(1ωt) and b1sin(1ωt)
It enables easy Tigonometric Fourier Series to Complex Fourier Series transformation.

Fig.3-3
a0-constant
a1,b1,a2,b2…-appropriate cosinus/sinus amplitudes
Are the a0,a1,b1,a2,b2…any value?—> The Fig.3-3 is a  common formula for cosinus/sinus sum with different a1,b1,a2,b2…amplitudes and 1ω, 2ω, 3ω… pulsations.
Are the a0,a1,b1,a2,b2… calculated acc. Fig. 2-12 Chapter 2.2 formula?—>The Fig.3-3 is a Tigonometric Fourier Series formula.
Let’s consider a1=1, b1=0.7 and ω=2.62/sec.
I.e. We are looking rotating vectors for:
f(t)=1cos(2.62t)+0.7sin(2.62t)
The vectors may be treated as complex numbers here.

Fig. 3-4
a1cos(1ωt) and b1sin(1ωt) components as vertical projection for real axis Re z
Fig. 3-4a
Rotating vectors z1z2 and the sum c1 intial state i.e. for 1ωt=0.
z1=a1=1
z2=-jb1=-j0.7
c1=z1+z2=1-j0.7
–>see*Note
Fig. 3-4b
Rotating vectors z1z2 and the sum c1 after time when 1ωt=π/3=60º
z1*exp(j1ωt)=a1*exp(jωt)=1*exp(j1ωt)
z2*exp(jωt)=-jb1*exp(1jωt)=-j0.7*exp(j1ωt)
c1=z1+z2=(a1-jb1)*exp(j1ωt)=(1-j0.7)*exp(j1ωt)
The crucial conclusion
The a1cos(1ωt)+b1sin(1ωt) is a rotating vector (a1-jb1)*exp(exp(j1ωt) vertical projection for real axis Re z
Or other words
 a1cos(1ωt)+b1sin(1ωt) is a real part of the complex function (a1-jb1)*exp(exp(j1ωt) 
Or other words
a1cos(1ωt)+b1sin(1ωt) =Re z{ (a1-jb1)*exp(j1ωt)}
The evidence is the Fig. 3-4b school trigonometry where blue and green vectors lengths are:
a1=1 and b1=0.7.
These rotating vectors represent the undermentioned f(t) functions.

Fig. 3-5
The f(t)=1.221cos(2.62t-35º)=1cos(2.62t)+0.7sin(2.62t) corresponding to rotating vectors from Fig-3-4       
*Note
c1=1-j0.7 
may be presented as module |c1|=1.221 and phase φ1≈-35º or equivalently as 1.221*exp(-j35º).
Module 1.221 is a and 0.7 Pitagoras and φ1=arctg(-0.7/1)≈-35º. Use calculator to check it.

Chapter 3.3 Complex Fourier Series for 3 harmonics
Chapter 3.3.1 Complex Fourier Series “with  real part Re”
We start with f(t) function with harmonics only. This function is easier to analyse.
f(t)=a1cos(1ωt)+b1sin(1ωt)+a2cos(2ωt)+b2sin(2ωt)+a3cos(3ωt)+b3sin(3ωt)
Some questions are coming up now.
First
Why do fragmentize the f(t) for Fourier harmonics? This function consists of  Fourier harmonics, after all!
You see a0, a1,b1,a2,b2 a3,b3 coefficients in f(t) and you don’t need to use the previous chapter Fig. 2-12 formula.
Yes, you are right. But the reason is didactical only. In any case, the formula may be used and the result is the same!
Second
Why “Fourier Series for 3 harmonics”? We see components in f(t), after all.
Because a1cos(1ωt)+b1sin(1ωt) may be treated as harmonic c1*sin(1ωt-φ1) .
Let’s f(t) is a particular function:
f(t)=1cos(1ωt)+0.2sin(1ωt)+0.6cos(2ωt)+0.4sin(2ωt)+0.4cos(3ωt)+0.4sin(3ωt)
i.e.
a1=1 b1=0.2
a2=0.6 b2=0.4
a3=0.4 b3=0.4
The basic pulsation ω may be any. But if you like nitty-gritty–>ω=2.62/sec as in animation Fig. 3-1.
c1, c2c3 are vectors or complex numbers:
c1=a1-jb1=1-0.2j
c2=a2-jb2=0.6-0.4j
c3=a3-jb3=0.4-0.4j
The f(t) function may be assign the rotating vectors or complex numbers now.
c1*exp(j1ωt)
c2*exp(j2ωt)
c3*exp(j3ωt)
The conclusion come from Fig. 3-4 now
a1cos(1ωt)+b1sin(1ωt)=Re z {(a1-jb1)*exp(j1ωt)} >i.e. real part of (a1-jb1)*exp(j1ωt)
Analogously for 2ωt,3ωt
a2cos(2ωt)+b2sin(2ωt)=Re z {(a2-jb2)*exp(j2ωt)}
a3cos(3ωt)+b3sin(3ωt)=
Re z {(a3-jb3)*exp(j3ωt)}
I.e.
f(t) time function is a real part of the rotating vectors sum
f(t)=c1*exp(j1ωt)+c2*expj2ωt)+c3*exp(j3ωt)
when c1=a1-jb1c2=a2-jb2c3=a3-jb3

Fig. 3-6
Fig. 3-6a Rotating vectors of the f(t) function
Fig. 3-6b f(t) function as a harmonics sum
c1c2c3  rotating vectors initial state when t=0
c1*exp(j1ωt)c2*exp(j2ωt)c3*exp(j3ωt) rotating vectors when 1ωt=30º (or 2ωt=60º or 3ωt=90º)
The “most delayed” vector c3 did the biggest rotation after time (when 3ωt=90º). No wonder, it has the biggest pulsation 3ωt!
The rotating vectors sum vertical projection  for Re z is just a f(t) time function!
I.e.
f(t)=Re z  {c1*exp(j1ωt)+c2*exp(j2ωt)+c3*exp(j3ωt)}
Two notes:
1.
 The rotating vectors returned to its initial state after T=2π/ω. But blue did 1 rotation, green 2 rotations and red rotations.
2. The vectors are similar to Solar system with Sun in center (0,0) and orbits:
c3*exp(j3ωt)
Venus orbit (the least)
c2*exp(j2ωt)Earth 
orbit
c1*exp(j1ωt)-Mars 
orbit  (the biggest)
Fig. 3-6 shows rotating vectors but you don’t see their sum!
Fig. 3-7 is without this fault. The rotating vectors sum is the end of the c3*exp(j3ωt).

Fig. 3-7
f(t)=1*exp(j1ωt) c2*exp(j2ωt) + c3*exp(j3ωt) as the end of red rotating vector
-The blue rotates around  (0,0)
-The green rotates around the  blue end
-The red rotates around the green end
It’s similar to the cracking whip when its end exceeds sound barrier.
Let’s return to astronomy.
Blue is an Earth rotating around Sun
Green is a Moon rotating around Earth
Red 
is a satellite rotating around Moon
Go for a moment to YouTube Fig. 3-18. All will be clear.
Conclusions

Fig. 3-8
Complex Fourier Series formula for f(t) with harmonics “with real part Re
Fig. 3-8a
General formula when any harmonics
Fig. 3-8b
Particular formula when 3 particular harmonics

Chapter 3.3.2 Complex Fourier Series “without real part Re”-other words “Oppositely rotating vectors version”
This is most popular Complex Fourier Series version. So, it is often and simply named Complex Fourier Series formula.

Fig. 3-9
Fig.3-9a 
and Fig.3-9b formulas are equivalent!
When we make changes in the formula Fig. 3-9a
1. Rotating (1-0.2j)*exp(j1ωt) vector will be replaced by the sum of
– it’s half –>(0.5-0.1j)*exp(j1ωt)
– it’s “conjugate half”–>(0.5+0.1j)*exp(-j1ωt). It rotates in the oppositely direction because there is “-j1ωt”
2. We do analogously with (0.6-0.4j)exp(j2ωt) and (0.4-0.4j)exp(j3ωt)
then
 the Fig. 3-9b formula is equivalent.
Note, that Fig. 3-9b formula length is double bigger, but the result is immadietaly real, without Re{…} operation!
Fig. 3-9c is a general formula.
Remember that c1, c2 and c3 are double smaller of the Fig. 3-8a c1, c2 and c3 coefficients!
Stricly-their modules-lengths are double smaller.

Fig. 3-10
f(t) as oppositely rotating vectors sum
– general formula for any c(-3)c(-2)c(-1)c1, c2 c3
– particular formula for concret c(-3)c(-2)c(-1)c1c2 c3
You see that:
1. Lower rotating vectors are halfs of the Fig.3-7 lower rotatting vectors  for t=0
2. Lower and higher vectors are conjugate complex numbers
3. Lower and higher vectors are contrary rotating and their sum is:
f(t)=a1cos(1ωt)+b1sin(1ωt)+a2cos(2ωt)+b2sin(2ωt)+a3cos(3ωt)+ b3sin(3ωt)
Note
 that there is no Re real part operand!–> compare with Fig 3-8
p.3 explanation
Formula shows contrary rotating vector pairs c(-1)c1  c(-2),c2  c(-3),c3  “snaphotted”  when t=0.
For example:
Contrary rotating vectors pair sum
(0.5-j0.1)exp(j1ωt) and (0.5+j0.1)exp(-j1ωt)
is a blue vector on Re z axis.
Conclusion 
(0.5-j0.1)exp(j1ωt)+(0.5+j0.1)exp(-j1ωt)=1cos(1ωt)+0.2sin(1ωt)
This same behaviour is for rotating pairs:
(0.3-j0.2)exp(j2ωt)+(0.3+j0.2)exp(-j2ωt)=0.6cos(1ωt)+0.4sin(1ωt)
(0.2-j0.2)exp(j3ωt)+(0.2+j0.2)exp(-j3ωt)=0.4cos(1ωt)+0.4sin(1ωt)
The Fig.3-9b particular formula is proved now. You can prove analogously Fig.3-9c general formula.

Chapter 3.4 Complex Fourier Series “with real part” formula
Chapter 3.4.1 Principles
We use the Chapter 2 knowledge here 
1.
 f(t) is any perodic function with period
2. 
Calculate  ω=2π/T pulsation
3. 
Calculate
– 
real  a0,a1,b1,a2,b2,…an,bn… and c0 coefficients
– 
complex c1,c2,c3…cn… coefficients

Fig. 3-11
Real and complex Fourier coefficients
a0,a1,b1,a2,b2…an, bn… are the same as Fig.2-12 Chapter 2.
When     c0, c1,c2,c3,c4…,cn… coefficients are calculated
then        f(t) real periodic function may be presented as:

Fig. 3-12
Complex Fourier Series-General formula with real part.
This is a Fig.3-8a formula  generalisation, where f(t) consists of harmonics only. The number of harmonics may be infinitive in Fig. 3-12! The formula approximates the f(t) function when n=∞ mostly. But finite n, for example n=20 approximates f(t) quite well. The a0=c0 component is constant. It doesn’t’t exist in Fig.3-8a.
Fig.3-13 shows the f(t) function components as rotating vectors.

Fig.3-13
The yellow vector c0  doesn’t rotate. Or formally rotates with ω=0. The others rotate “each around previous” end. I.e. around c0 constant indirectly. The next vector rotation speed is increasing. The consecutive vectors are smaller and the “final” vector c∞exp(j∞t) aims at mostly. The vertical projection of the “final” vector moves in Re z axis acc. to f(t) function!
Note
The “final” rotating vector may be not c∞exp(j∞t), but for example  c20*exp(j20t) in practice. The f(t) function will be quite well approximated then.

Chapter 3.4.2 Square wave example
The square wave was exactly analyzed as Trigonometric Fourier Series in Chapter 2.8. Glance at it for a while.
The square wave with ω=2π (i.e. T=1sec) is approximated as f(t)

Fig. 3-14
Fourier Trigonometric Series
 for square wave with ω=2π 1/sec.
The coefficients are calculated acc. Fig.3-11 
a0,a1a2a3,a4…=0
b2,b4,b6…=0
b1=4/1π, b3=4/3π, b5=4/5π,b7=4/7π
and the f(t) is as Fig. 3-14
The complex c1c3c5c7 coefficients are calculated then :
c1=a1-jb1=0-4j/π=-4j/π
c3=a3-jb3=0-4j/3π=-4j/3π
c5=a5-jb5=0-4j/5π=-4j/5π
c7=-4j/7π =0-4j/7π=-4j/7π
The Trigonometric Fourier Series of the Square Wave may be presented as Complex Fourier Series or as rotating vectors.

Fig. 3-15
Complex Fourier Series 
for Square Wave
The c1,c3,c5,c7 i.e. 4 first nonzero coefficients are used only, so the Square Wave  approximation will be not sensational.
You have to imagine the next c9,c11,c13,…c∞ vectors.
There are presented as c1exp(j1ωt),c3exp(j3ωt),c5exp(j5ωt),c7exp(j7ωt) rotating vectors in 2 states:
when 1ωt=0° (initial state)
when 1ωt=60° (or 3ωt=180° or 5ωt=300° or 7ωt=420°)
The fastest is the c7exp(j7ωt) vector and it made the biggest rotation=420°–>full rotation 360°+60°.
The vertical end of the c7exp(j7ωt) vector moves as f(t) square wave   approximation–>. Why approximation and not ideal square wave? Because it wasn’t “infinte” as c∞exp(j∞ωt) vector. The more vectors are used, the f(t) is more similar to the ideal square wave.
Where is the hypothetical c∞exp(j∞ωt) end position? It lays somewhere near c7exp(j7ωt) vector end.
All will be clear after animations.

Chapter 3.4.3 Animations
Animation 1

Fig. 3-16
You see the Fig. 3-15 version but with the trajectories added. I.e The first is a rotating c1*exp(jωt) vector, the second-c3*exp(j3ωt) and the last-c5*exp(j5ωt). The c7*exp(j7ωt) from Fig. 3-15 doesn’t exist. It isn’t necessary to understand the idea.
I recommend You the wonderful site.  Pierre Guilleminot is an author–>author
You can change the f(t) function and the harmonics number. The bigger is this number, the f(t) approximation is more accurate. Convince Yourself! You can test different functions square wave, triangel,sawtooth, pulse….

Fig. 3-17
Square wave simulation
Click ANIMATION
The picture is clockwise rotated when compared to Fig. 3-16.
Select with the aid of mouse Square function, harmonics and not too big simulation velocity.
The picture is coming alive! You will be convinced now that that Fourier Series Fig. 3-12 is similar to the Solar System.
Solar is the (0,0), Earth is the end of c1*exp(j1ωt) vector, Moon-c3*exp(j3ωt) and Moon satellite-c5*exp(j5ωt).
The end of the  c5*exp(j5ωt) vector moves strictly to the formula:
f(t)=Re
{c1*exp(j1ωt)+c3*exp(j3ωt)+c5*exp(j5ωt)}
I propose you to test other functions-triangel, sawtooth…

Animation 2-YouTube
Click YouTube triangle. Author’s Nick–>GLV

Fig.3-18
Complex Fourier Series Animation of the square wave
The 4 upper rotating vectors correspond to the 4 lower formula components.
f(t)=Re{c1exp(j1ωt)+c3exp(j3ωt)+c5exp(j5ωt)+c7exp(j7ωt)}
Do You see the Solar System in the lower rotating circles?
Sun  is a (0,0) stationary point
The blue circle is an Earth rotating around the Sun  corresponding to the c1exp(j1ωt).
The green circle is a Moon rotating around the Earth–>c3exp(3ωt)
The red circle is a satelite  rotating around the Moon –>c5exp(5ωt)
The beetroot circle is a satelite of  the Moon satellite –>c7exp(j7ωt)
In fact, the last “satelite squared”  doesn’t exist , but please imagine yourself.

Chapter 3.5 Complex Fourier Series-Ultimate formula “without real part”
1. There is a periodical f(t) function with a period T.
2. Calculate ω=2π/T primary pulsation.
3. Calculate a0,a1,b1,a2,b2,…an,bn… real coefficients and then … c(-n)…c(-3),c(-2),c(-1),c0,  c1,c2,c3…cn… complex coefficients.
You have Complex Fourier Series now.

Fig. 3-19
Complex Fourier Series coefficients-Ultimate formula
The formula in comparison to Fig. 3-11
1. c0 is identical
2. an, bn  are the halfs of  an, bn from the Fig. 3-11
3. c(n) and c(-n) pairs are conjugate numbers.
Fig. 3-20
Complex Fourier Series -Ultimate formula
The negative c(n) coefficients emerged. Their number is doubled and their modules  are double smaller than in Fig. 3-12 (except c0).
But their results  are the same  f(t) real functions!

Fig. 3-21
Complex Fourier Series 
as a sum of the oppositely rotating vectors.
More strictly-sum+c0 coefficient.
Justification
Every oppositely rotating vectors pair-for example:
c2exp(j2ωt)+c(-2)exp(-j2ωt)=a2cos(2ωt)+ b2sin(2ωt)–> Chapter 3.3.2.
I.e. all the pairs sum+c0  (c0=a0 ) gives the Fig. 3-3 i.e. Fig. 3-19 formula
The most elegant and condensed formula form:

Fig. 3-22
Complex Fourier Series
This is formula “without real part” and more frequently used than “with real part”–>Fig. 3-12. So this is simply Complex Fourier Series

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